This question popped up to me a few moments ago. I was measuring what is intended to be a 50Mhz square wave of level 0 to 2.5, however what I saw on the screen is a sine wave that was centered around 1.2V and level of 0.5 to 2.0V, the frequency was 4MHz.

I checked up my oscilloscopes datasheet and it showed that the bandwidth was 10MHz with a sampling rate of 50 MS/s.

I'm wondering what these figures are all about.

  • Are they a measure of the upper frequency limit an oscilloscope can measure?
  • Is this oscilloscope capable of measuring 50Mhz at all?
  • \$\begingroup\$ Can you make a measurement on a sine wave rather than a square wave? What you call a 50MHz square wave is really a combination of higher-frequency sine waves. \$\endgroup\$ Commented Jan 5, 2011 at 4:10
  • \$\begingroup\$ Nice idea! I don't have such a high frequency source though, unless off course I build one? \$\endgroup\$
    – Kevin Boyd
    Commented Jan 5, 2011 at 8:15
  • \$\begingroup\$ Sure, you could build one with a narrow bandpass filter (opamp and some caps/resistors) centered at your desired frequency, and drive it with your square wave. National Semi has an article on that, here's a sine and cosine generator, and here's the Falstad simulation of that generator. \$\endgroup\$ Commented Jan 5, 2011 at 19:34

4 Answers 4


System bandwidth is a combination of probe bandwidth and oscilloscope input bandwidth. Each can be approximated by an RC lowpass circuit, which means delays add geometrically:

t_system^2 = (t_probe^2 + t_scope^2)
f_system = 1/sqrt((1/f_probe)^2 + (1/f_scope)^2)

This means that a 10MHz 'scope with 60MHz probes can measure sinusoids of frequency 9.86MHz with -3dB (100*10^{-3/20}%) attenuation.

When measuring digital pulse trains it's not so much the periodicity that matters, but the rise and fall times, as they contain the high-frequency information. Rise times can be approximated mathematically by an RC rise or a Gaussian rise, and are defined as the time for the signal to go from 10% of the difference between low voltage (logical 0) and high voltage (logical 1), to 90% of the difference. For example, in a 5V/0V system, it is defined as the time to get from 0.1*5V=0.5V to 0.9*5V=4.5V. With these constraints and some fancy math, one can work out that each type of characteristic rise time has frequency content up to about 0.34/t_rise for Gaussian and 0.35/t_rise for RC. (I use 0.35/t_rise for no good reason and will do so for the rest of this answer.)

This information works the other way, too: a particular system bandwidth is only able to measure rise times up to 0.35/f_system; in your case, 35 to 40 nanoseconds. You're seeing something similar to a sine wave because that is what the analog front-end is letting through.

Aliasing is a digital sampling artifact, and is also in effect in your measurement (aren't you lucky!). Here's a borrowed image from WP:

Aliasing example.

As the analog front-end is only letting rise times 35ns to 40ns through, the ADC sampling bridge sees something like an attenuated 50MHz sine wave, but it's only sampling at 50MS/s, so it can only read sinusoids below 25MHz. Many 'scopes have an antialiasing filter (LPF) at this point, which would attenuate frequencies above 0.5 times the sample rate (Shannon-Nyquist sampling criteria). Your scope doesn't seem to have this filter, though, as the peak-to-peak voltage is still fairly high. What model is it?

After the sampling bridge the data gets shoved into a few DSP processes, one of which is called decimation and cardinal spans, which further reduces sample rate and bandwidths in order to better display and analyze it (especially helpful for FFT calculation). The data is further massaged such that it doesn't display frequencies above ~0.4 times the sample rate, called a guard band. I would have expected you to see a ~20MHz sinusoid -- do you have averaging (5-point) turned on?

EDIT: I'll stick my neck out and guess that your oscilloscope has digital antialiasing, using decimation and cardinal spans, which basically means a digital LPF then resampling of an interpolated path. The DSP program sees a 20MHz signal, so it decimates it until it is below 10MHz. Why 4MHz and not closer to 10MHz? "Cardinal span" means halving the bandwidth, and decimation is often by a power of two as well. Some integer power of 2 or a simple fraction of it resulted in a 4MHz sinusoid being spat out instead of ~20MHz. This is why I say every enthusiast needs an analog 'scope. :)

EDIT2: Since this is getting so many views, I'd better correct the above embarrassingly thin conclusion.
EDIT2: The particular tool you liked to can use undersampling, for which a windowing analog BPF input is required for antialiasing, which this tool doesn't seem to have, so it must only have a LPF, restricting it to sinusoids of less than 25MHz even when using equiv. time sampling. Although I also suspect the quality of the analog side, the digital side likely does not do the aforementioned DSP algorithms, instead streaming data or transferring one capture at a time for brute force number crunching on a PC. 50MS/s and 8-bit word lengths means this is generating ~48MB/s of raw data -- far too much to stream over USB despite its theoretical 60MB/s limit (practical limit is 30MB/s-40MB/s), nevermind the packetizing overhead, so there is some decimation right out of the box to reduce this. Working with 35MB/s gives ~37MS/s sample rate, pointing to a theoretical measurement limit of 18MHz, or 20ns rise time, when streaming, though it is likely lower as 35MB/s is amazing (but possible!). The manual indicates a Block Mode exists for capturing data at 50MB/s 'til the internal 8k memory (cough) is full (160us), then sending it to the computer at a leisurely pace. I would assume that the difficulties encountered in designing a quality analog input were partially overcome by oversampling by 2X (extra half-bit accuracy), giving an effective sample rate of 25MS/s, maximum frequency 12.5MHz, and a 10% guard band ((0.5*25-10)/25), all of which could be reduced in the hand-tool itself. In conclusion, I'm not sure why you're seeing a 4MHz sinusoid as there are ways for this to happen, but would want to make the same measurement in Block Mode then analyze the data with a third-party program. I have always been hard on PC-based oscilloscopes, but this one seems to have decent inputs...

  • \$\begingroup\$ Note that some digital systems (CMOS in particular) define rise times from 20% to 80%. \$\endgroup\$
    – tyblu
    Commented Jan 4, 2011 at 9:05
  • \$\begingroup\$ Upvote!! quite technical and somethings that I don't understand!! ;) \$\endgroup\$
    – Kevin Boyd
    Commented Jan 4, 2011 at 9:36
  • \$\begingroup\$ The scope is here picotech.com/handheld-oscilloscope.html, model 2104 \$\endgroup\$
    – Kevin Boyd
    Commented Jan 4, 2011 at 9:39
  • \$\begingroup\$ Any recommendations for 2 channel scope for 100Mhz? \$\endgroup\$
    – Kevin Boyd
    Commented Jan 4, 2011 at 9:49
  • 1
    \$\begingroup\$ @KevinBoyd, Right now the Instek GDS-1062A looks quite nice with 2M memory and 60MHz. I have a Rigol DS1052E with 1M memory, and may bump it up to 100MHz, but that isn't as important to me as the word memory. For higher frequencies I got an old Tek 485: 4 channels, 200MHz. I think Rigol has changed their firmware recently to stop the hack. Info: A, B. \$\endgroup\$
    – tyblu
    Commented Jan 4, 2011 at 10:59

10 MHz analog bandwidth means that a 10 MHz signal at 10 V will look like a 5 V, in other words your amplitude will be halved at 10 MHz.

The 10 MHz bandwidth means that your 50 MHz signal will be dampened down quite a bit, but how much is hard to speculate on.

The 50 MS/s means that you cannot realistically work with signals much over 5 MHz if you hope to do single-shot capture of the signal, which is really the only reason to have a DSO in the first place.

Ignoring the bandwidth problem for a minute, you might be able to put the scope into repetitive sampling mode and capture a repetitive signal that way, just like an analog scope does.

I'd get a proper DSO (The Rigol ds1052e modded to 100 MHz analog bandwidth gets my recommendations), failing that a used Tektronix analog scope might be a good way to go (I use the 2236, 2246 and 2247A models from time to time and they are all fine analog scopes)

  • 5
    \$\begingroup\$ 10 MHz would be the point of 3dB attenuation, so 0.707x not 0.5x. \$\endgroup\$
    – Thomas O
    Commented Jan 4, 2011 at 8:45
  • \$\begingroup\$ @dren.dk: This is a pc based oscilloscope. For measuring digital signals upto 100Mhz, what scope and probes would you recommend? \$\endgroup\$
    – Kevin Boyd
    Commented Jan 4, 2011 at 9:41
  • 1
    \$\begingroup\$ @Kevin Boyd youtube.com/watch?v=LnhXfVYWYXE \$\endgroup\$
    – Thomas O
    Commented Jan 4, 2011 at 10:43
  • 1
    \$\begingroup\$ @Kevin: There is no problem modding the Rigol as you can still downgrade to the older firmware, do the mod, then upgrade to the latest and greatest. They will need to change the actual hardware to get rid of the mod. \$\endgroup\$
    – dren.dk
    Commented Jan 4, 2011 at 12:59
  • 1
    \$\begingroup\$ @dren.dk, You are correct - I see the fix here. Software version 02.05 SP1 ( seems to have a trigger bug, so stay at 2.04 SP1. \$\endgroup\$
    – tyblu
    Commented Jan 4, 2011 at 19:29
* Are they a measure of the upper frequency limit an oscilloscope can measure?

Yes for direct measurment.

* Is this oscilloscope capable of measuring 50Mhz at all?

Yes, using some tricky ways: 1) peak detection (useful when you need to see AM modulated signal) 2) Freq shift (again, useful when signal is modulated) - If you mix 50Mhz signal with 49Mhz sine wave, you'll get 1Mhz signal near the frequency you want.

  • \$\begingroup\$ What is direct measurement? \$\endgroup\$
    – Kevin Boyd
    Commented Jan 4, 2011 at 9:50
  • \$\begingroup\$ When you just connect the scope to your signal and want to see everything. \$\endgroup\$ Commented Jan 4, 2011 at 10:00

The bandwidth and the sample rate normally should be 4 to 5 times the maximum frequency you want to measure. But be aware that if your input signal is not a pure sine wave, as in your case the square wave, it also contains harmonics with much higher frequencies. For an accurate measurement you have to cover at least the first of these harmonics.

At the frequency of the maximum bandwidth (here 10 MHz) a sine wave of this frequency is attenuated by 3dB by the analog frontend of the scope. This means it is measured at only 70% of its real value. The sample rate specifies how many measurements are done by the scope per second i.e. how accurate the form of the signal is acquired (50 MS/s equals 5 measurements per cycle on a 10 MHz signal).

Now think about what your scope is seeing with the input signal heavy attenuated (due to the too low bandwith) and with only 5 samples per cycle (due to the sampling rate).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.